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Context: This question arose as I was reading the proof of Application 6.1 in Mumford's Abelian Varieties. However, I have extracted all of the relevant information below so this question should stand alone.

Let $X$ be a complete variety over an algebraically closed field $k$. Let $D$ be an effective divisor on $X$ such that the linear system $|D|$ is without basepoints. (That is, there is no point $P \in X$ such that $P$ is in the support of every divisor on $X$ that is linearly equivalent to $D$.) Let $L$ be the invertible sheaf associated to the divisor $D$. Then $|D|$ gives rise to a morphism $\phi : X \to \mathbb{P}^N$, where $N + 1 = \dim \Gamma(X, L)$. (I will describe this morphism in more detail below.) If $\phi$ is not a finite morphism, we can find an irreducible curve $C$ in $X$ such that $\phi(C)$ is a single point. Mumford claims that if $E$ is any effective divisor on $X$ that is linearly equivalent to $D$ then either $E$ contains $C$ or $E$ is disjoint from $C$. My question is: why is this true? Why can't $E$ intersect $C$ in a finite set of points?

Construction of $\phi$: Choose a basis $s_0, \ldots, s_N \in \Gamma(X, L)$. Let $X_i$ be the set of $x \in X$ such that $s_i \not\in \mathfrak{m}_xL_x$. Let $U_i = \mathrm{Spec}\,k[y_0, \ldots, \hat{y}_i, \ldots, y_N]$. Define a ring homomorphism $f_i : k[y_0, \ldots, \hat{y}_i, \ldots, y_N] \to \Gamma(X_i, \mathcal{O}_X)$ by $f(y_j) = s_j/s_i$. Then by composing the map induced by $f_i$ on spectra with the natural map $\alpha_i : X_i \to \mathrm{Spec}\, \Gamma(X_i, \mathcal{O}_X)$ we obtain $\phi_i : X_i \to \mathrm{Spec}\, k[y_0, \ldots, \hat{y}_i, \ldots, y_N]$. These maps glue together to define our morphism $\phi : X \to \mathbb{P}^N$.

Small reduction: Changing $\phi$ by an automorphism of $\mathbb{P}^N$ corresponds to choosing a different basis for $\Gamma(X, L)$. By doing this we may assume that $\phi(C) = [1 : 0 : \cdots : 0]$. Thus $\phi(C) \in U_0$ and $\phi(C) \not\in U_i$ for $1 \leq i \leq N$. Thus $C \subset X_0$. Now, divisors on $X$ give rise to divisors on $X_0$ by restriction, so we can reduce our problem to showing that $C$ is either contained in or disjoint from any divisor of $X_0$ that is linearly equivalent to the restriction of $D$. The apparent advantage of this (though I haven't been able to make good use of it) is that the map $\phi_0$ is fairly explicit and does not involve glueing morphisms together.

One last comment: We can identify $\mathbb{P}^N$ with $|D_0|$ via $[a_0 : \cdots : a_N] \leftrightarrow \sum_{i = 0}^N a_i (s_i)_0$, where $(s_i)_0$ is the divisor of zeros of the global section $s_i$. Under this interpretation, $\phi(C)$ is an effective divisor linearly equivalent to $D$. Perhaps this is a useful way to approach the question?

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Saying that $E$ is linearly equivalent to $D$ means that you can find a hyperplane $H$ in $\mathbb{P}^N$ such that $E = f^*H$. Now, either $H$ contains the point $\phi(C)$, or it doesn't. – Vesselin Dimitrov Feb 3 '13 at 0:22
I mean, $E = \phi^* H$. – Vesselin Dimitrov Feb 3 '13 at 0:24
A minor correction regarding your last comment: the $\mathbb P^N$ which is the image of $\phi$ should be identified with the dual of $\lvert D_0\rvert$. Of course, these are both projective spaces of the same dimension, but they are not canonically isomorphic. – Dustin Cartwright Feb 3 '13 at 15:30
Why is the dual space more appropriate? – anonymous Feb 3 '13 at 19:17
up vote 2 down vote accepted

$E$ can't intersect $C$ in a finite number of points because otherwise the restriction of $\phi$ to $C$ would be a finite degree morphism, which you assume is not. $E$ is the pull-back of the hyperplane divisor, as Vesselin remarked.

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I am sorry if this is pedantic, but I do not see why the definition of $\phi$ makes $E$ the pull-back of a hyperplane in $\mathbb{P}^N$. – anonymous Feb 3 '13 at 19:12
By def, $E$ is lin. eq. to the pull-back of the intersection of an hyperplane with the image of $X$. This is what is - with a little abuse of terminology - known as hyperplane divisor. – IMeasy Feb 4 '13 at 7:20

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